Hi Abdullah,
Although not stated explicitly in the Question, I'm going to assume that
a) the objective is compute the convolution integral, i.e., xt and ht are continuous time signals, and
b) xt and ht are both zero for t < 0 adn t > 10
If those assumptions are correct, here's a similar problem that shows how to use conv to approximate the convolution integral. Maybe you can adapt it to your problem. Let's use two simple functions for x(t) and h(t), both of which are zero outside the interval 0 <= t <= 3
x(t) = exp(-0.2*t)*rectangularPulse(0,3,t);
h(t) = exp(-0.3*t)*rectangularPulse(0,3,t);
By defintion, the convolution integral for these signals is
y(t) = simplify(int(x(tau)*h(t-tau),tau,0,t))
Plot x(t), h(t), and y(t). The duration of y(t) (6 seconds) is the sum of the durations of x(t) and h(t).
fplot([x(t) h(t) y(t)],[0 7])
Now use conv with numerical valuse of x(t) and h(t) sampled at 0.01 seconds
To get the correct answer, we need to use the full convolution and multiply the result by the sampling period (because both x(t) and h(t) are finite)
y = conv(x,h,'full')*0.01;
Of course, y will be longer than x and h, so we have to define a new time vector to plot it.
plot((0:numel(y)-1)*0.01,y)
For this problem, if we use the 'same' convolution, we get a y that has the same number of elements x and h and therefore covers 3 seconds. But that that's not sufficient to approximate the 6-second convolution integral. And we can see that plotting it would require a shift to the right by 1.5 seconds to match up with the "central" part of the convolution integral.
y = conv(x,h,'same')*0.01;
legend('x(t)','h(t)','y(t)','yconv-full','yconv-same')
